The pressure at what depth in seawater (density 1,025 kg/m3) is equivalent to that at a depth of 10.0 m in fresh water (density 1,000 kg/m3)?

To solve this problem, we need to use the formula for pressure due to a fluid column, which is P = ρgh, where P is the pressure, ρ is the fluid density, g is the acceleration due to gravity, and h is the height (or depth in this case) of the fluid column.

1. First, calculate the pressure at a depth of 10.0 m in fresh water. We know that the density of fresh water (ρ1) is 1,000 kg/m3, g is approximately 9.81 m/s2, and h1 is 10.0 m. So, the pressure P1 is:

   P1 = ρ1 * g * h1 P1 = 1,000 kg/m3 * 9.81 m/s2 * 10.0 m P1 = 98,100 Pa

2. Next, we want to find the depth in seawater that would give us the same pressure. We know that the density of seawater (ρ2) is 1,025 kg/m3. We can set up the equation P1 = P2, where P2 = ρ2 * g * h2, and solve for h2:

   98,100 Pa = 1,025 kg/m3 * 9.81 m/s2 * h2

3. Solve for h2:

   h2 = 98,100 Pa / (1,025 kg/m3 * 9.81 m/s2) h2 = 9.76 m

So, the pressure at a depth of 9.76 m in seawater is equivalent to the pressure at a depth of 10.0 m in fresh water.